Hdu 4965（矩阵快速幂）

题目链接

Fast Matrix Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 87    Accepted Submission(s): 39

Problem Description

One day, Alice and Bob felt bored again, Bob knows Alice is a girl who loves math and is just learning something about matrix, so he decided to make a crazy problem for her.

Bob has a six-faced dice which has numbers 0, 1, 2, 3, 4 and 5 on each face. At first, he will choose a number N (4 <= N <= 1000), and for N times, he keeps throwing his dice for K times (2 <=K <= 6) and writes down its number on the top face to make an N*K matrix A, in which each element is not less than 0 and not greater than 5. Then he does similar thing again with a bit difference: he keeps throwing his dice for N times and each time repeat it for K times to write down a K*N matrix B, in which each element is not less than 0 and not greater than 5. With the two matrix A and B formed, Alice’s task is to perform the following 4-step calculation.

Step 1: Calculate a new N*N matrix C = A*B.
Step 2: Calculate M = C^(N*N). 
Step 3: For each element x in M, calculate x % 6. All the remainders form a new matrix M’.
Step 4: Calculate the sum of all the elements in M’.

Bob just made this problem for kidding but he sees Alice taking it serious, so he also wonders what the answer is. And then Bob turn to you for help because he is not good at math.

Input

The input contains several test cases. Each test case starts with two integer N and K, indicating the numbers N and K described above. Then N lines follow, and each line has K integers between 0 and 5, representing matrix A. Then K lines follow, and each line has N integers between 0 and 5, representing matrix B.

The end of input is indicated by N = K = 0.

Output

For each case, output the sum of all the elements in M’ in a line.

Sample Input

4 2
5 5
4 4
5 4
0 0
4 2 5 5
1 3 1 5
6 3
1 2 3
0 3 0
2 3 4
4 3 2
2 5 5
0 5 0
3 4 5 1 1 0
5 3 2 3 3 2
3 1 5 4 5 2
0 0

Sample Output

14
56

思路：

C = A * B，　M = C^(n*n)，这题我的第一反应就是矩阵乘法啊。。但是却直接把A * B求了出来，然后计算Ｃ的ｎ×ｎ次幂

而这题解题的关键就在这里，因为Ａ是ｎ×ｋ的矩阵，Ｂ是ｋ×ｎ的矩阵，而ｎ很大，ｋ很小。

所以可以把Ｍ写成　Ｍ　＝　A*(B*A)*(B*A)*....*(B*A)*B

这样就这样用快速幂求得（Ｂ×Ａ）^(n*n-1)，然后右乘一个Ｂ，再左乘一个Ａ即可。智商真是拙计。！！

Accepted Code:

 /*************************************************************************
     > File Name: 1006.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年08月19日 星期二 12时05分28秒
     > Propose:
  ************************************************************************/

 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <fstream>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 struct mat {
   int n, m;
   vector<vector<int> >M;
   mat() {}
   mat(int a, int b):n(a), m(b) {
         M.resize(n);
         for (int i = ; i < n; i++) M[i].resize(m, );
   }
   friend mat operator * (const mat &a, const mat &b) {
      mat c(a.n, b.m);
      for (int i = ; i < a.n; i++) {
          for (int k = ; k < b.n; k++) {
              for (int j = ; j < b.m; j++) {
                  c.M[i][j] += a.M[i][k] * b.M[k][j];      //注意这里的循环顺序
                  c.M[i][j] %= ;
              }
          }
      }
      return c;
   }
 };
 void read(int &res) {
       res = ; char c = ' ';
     while (c < '' || c > '') c = getchar();
     while (c >= ''&&c<='') res = res*+c-'', c = getchar();
 }

 int main(void) {
       int n, k;
     while (~scanf("%d %d", &n, &k) && n + k) {
           mat A(n, k), B(k, n);
           for (int i = ; i < n; i++) for (int j = ; j < k; j++)
               read(A.M[i][j]);
           for (int i = ; i < k; i++) for (int j = ; j < n; j++)
               read(B.M[i][j]);
         mat C = B * A;
         int b = n * n - ;
         mat res(k, k);
         for (int i = ; i < k; i++) res.M[i][i] = ;
         while (b) {
               if (b&) res = res * C;
             C = C * C;
             b >>= ;
         }
         mat ans = A * (res * B);
         int sum = ;
         for (int i = ; i < n; i++) for (int j = ; j < n; j++)
               sum += ans.M[i][j];
         printf("%d\n", sum);
     }

     return ;
 }